GATE 2026 PH – Question 36

Correct answer: (B) $2i$ and $-2i$

Explanation

Factorise the numerator and the denominator.

**Numerator:** $z^2-5z+4=(z-1)(z-4)$.

**Denominator:** $z^3+4z-z^2-4=z^3-z^2+4z-4=z^2(z-1)+4(z-1)=(z-1)(z^2+4)$.

So
$$f(z)=\frac{(z-1)(z-4)}{(z-1)(z^2+4)}=\frac{z-4}{z^2+4}\quad(z\neq1).$$

• At $z=1$ the common factor cancels, so $z=1$ is a **removable singularity**, and the function can be defined there by continuity.
• The remaining zeros of the denominator, $z^2+4=0\Rightarrow z=\pm2i$, are **genuine (pole) singularities**.

Answer **$z=2i$ and $z=-2i$** (option B).